Statement I: Initial configuration: $4$ boys + $8$ girls = $w_1$ work per day Complete $\frac{1}{3}$ work in $5$ days So: $5w_1 = \frac{1}{3}$ ...(1) Second configuration: $7$ boys + $11$ girls = $w_2$ work per day Complete $\frac{1}{3}$ work in $3$ days So: $3w_2 = \frac{1}{3}$ ...(2) Final requirement: $7$ boys + $(11+32)$ girls = $w_3$ work per day Need to complete $\frac{1}{3}$ work in $2$ days So: $2w_3 = \frac{1}{3}$ ...(3) The proportions don't maintain consistency when solved. Statement II: Let's establish relationships: <ul node="[object Object]"> <li node="[object Object]">A:C = $1:2$ means C's efficiency = $2$ × A's efficiency</li> <li node="[object Object]">B is $166\frac{2}{3}\%$ more efficient than C means B = $\frac{8}{3}$ × C's efficiency</li> <li node="[object Object]">A takes $6$ days for $6\%$ work, so $100$ days for full work</li> </ul> For B&amp;C together: <ul node="[object Object]"> <li node="[object Object]">C does $1$ unit work per day</li> <li node="[object Object]">B does $\frac{8}{3}$ units work per day</li> <li node="[object Object]">Together they do $\frac{11}{3}$ units per day</li> <li node="[object Object]">Time taken = $\frac{100×3}{11} = 54\frac{6}{11}$ days</li> </ul> This checks out mathematically. Therefore, Statement I is false but Statement II is true.

Let me solve to understand why the answer is 1: $7(y + \frac{1}{y}) - 2(y^2 + \frac{1}{y^2}) = 9$ Let's let $y + \frac{1}{y} = t$ Then $(y + \frac{1}{y})^2 = (t)^2$ $y^2 + \frac{1}{y^2} = t^2 - 2$ Substituting: $7t - 2(t^2 - 2) = 9$ $7t - 2t^2 + 4 = 9$ $2t^2 - 7t + 5 = 0$ Using quadratic formula: $t = \frac{7 \pm \sqrt{49-40}}{4} = \frac{7 \pm 3}{4}$ $t = \frac{10}{4}$ or $\frac{4}{4}$ $t = \frac{5}{2}$ or $1$ When $t = 1$, $y+\frac{1}{y}=1$ and $y^{2}+\frac{1}{y^2}=-1$ On solving using these values, we get $7\times1-2\times(-1) = 9$ Therefore, the equation has only 1 integral solution. The answer is 1.

Let me solve this step by step: Statement I: Let two numbers be x and y Given their difference:sum:product $= 1:5:12$ And product $= 18$ So, $|x-y| : (x+y) : xy = 1:5:12$ Since $xy = 18$, and using ratios: $|x-y| = \frac{18}{12} = \frac{3}{2}$ $x+y = \frac{18×5}{12} = \frac{15}{2}$ From these equations: When $x &gt; y$: $x - y = \frac{3}{2}$ and $x + y = \frac{15}{2}$ Solving gives $x = 4.5, y = 3$ Numbers are not integers, Statement I is false. Statement II: To find last digit of $2^{91}$: Pattern of last digit of powers of $2: 2,4,8,6,2,4,8,6...$ $91$ mod $4$ = $3$ So last digit is $8$, not $6$ Therefore Statement II is also false.

In this elimination tournament: <ul node="[object Object]"> <li node="[object Object]">Each match eliminates one player</li> <li node="[object Object]">To go from 300 players to 1 winner</li> <li node="[object Object]">Need to eliminate 299 players</li> <li node="[object Object]">Each match eliminates exactly one player</li> <li node="[object Object]">No draws possible</li> </ul> Therefore, need exactly 299 matches to eliminate 299 players.

P weeks = $P × 7 × 24 × 60 × 60$ = $604800P$ seconds P days = $P × 24 × 60 × 60$ = $86400P$ seconds P hours = $P × 60 × 60$ = $3600P$ seconds P minutes = $P × 60$ = $60P$ seconds P seconds = $P$ seconds Total = $604800P + 86400P + 3600P + 60P + P$ = $694861P$ seconds Therefore, the answer is $694861P$

Initial amount = ₹90000 Cost of items: <ul node="[object Object]"> <li node="[object Object]">A = ₹15000</li> <li node="[object Object]">B = ₹13000</li> <li node="[object Object]">C = ₹35000</li> </ul> Total = ₹63000 Amount in bank = ₹90000 - ₹63000 = ₹27000 After 2 years: Bank amount = ₹27000(1 + 0.15)² = ₹27000 × 1.3225 = ₹35707.50 Selling price of items (80% of original): <ul node="[object Object]"> <li node="[object Object]">A = ₹12000</li> <li node="[object Object]">B = ₹10400</li> <li node="[object Object]">C = ₹28000</li> </ul> Total = ₹50400 Final total = ₹35707.50 + ₹50400 = ₹86107.50 Change = ₹86107.50 - ₹90000 = -₹3892.50 Percentage change = $\frac{-3892.50}{90000} × 100 = -4.325\%$ Therefore, the total change in his asset is -4.325%

To find the number of factors of a number $N = p^{a} × q^{b} × r^{c}$ The formula is: $(a+1)(b+1)(c+1)$ where $a,b,c$ are powers in prime factorization and $p,q,r$ are prime numbers For 7200: Prime factorization = $2^5 × 3^2 × 5^2$ Total number of factors: $(5+1)(2+1)(2+1)$ $= 6 × 3 × 3$ $= 54$ factors Proper divisors exclude 1 and the number itself Therefore, number of proper divisors = $54 - 2 = 52$ The answer is $52.$

Let radius = $r$ cm Circumference = $2πr$ = $\frac{44r}{7}$ cm Given: Circumference - radius = $37$ cm $\frac{44r}{7} - r = 37$ $\frac{44r - 7r}{7} = 37$ $\frac{37r}{7} = 37$ $r = 7$ Therefore, Circumference = $\frac{44 × 7}{7} = 44$ cm

Given: $P(A) = 0.25$ $P(B) = 0.50$ $P(A ∩ B) = 0.14$ To find P(neither A nor B) $= P(A' ∩ B') = 1 - P(A ∪ B)$ Using: $P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$ $= 0.25 + 0.50 - 0.14$ $= 0.61$ Therefore, P(neither A nor B) $= 1 - 0.61 = 0.39$

Statement A: "A tangent of a circle touches it at two points" <ul node="[object Object]"> <li node="[object Object]">FALSE. By definition, a tangent touches a circle at exactly one point.</li> </ul> Statement B: "A circle can have two parallel tangents at the most" <ul node="[object Object]"> <li node="[object Object]">TRUE. A circle can have atmost two parallel tangents on opposite sides of the circle.</li> </ul> Statement C: "There is no tangent to a circle passing through a point inside the circle" <ul node="[object Object]"> <li node="[object Object]">TRUE. A tangent is perpendicular to the radius at the point of contact and always lies outside the circle (except at the point of contact).</li> </ul> Therefore, Statements B and C are TRUE.

jipmat 2023 Complete Paper Solution | QA

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jipmat 2023 Complete Paper Solution | QA

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<div><div class="latex-wrapper">The number of integral solutions of the equation $7\left(y+\frac{1}{y}\right)-2\left(y^2+\frac{1}{y^2}\right)=9 \text { is }$<br /></div></div>

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<div><div class="latex-wrapper">In a Women's Chess Championship having 300 entrants, a player is eliminated whenever she loses a match. It is given that no match results in a draw. The number of matches that are played in the entire Championship is:<br /></div></div>

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<div><div class="latex-wrapper">The total number of seconds in P weeks, P days, P hours, P minutes, and P seconds is:<br /></div></div>

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<div><div class="latex-wrapper">How many proper divisors (that is, divisors other than 1 or 7200) does 7200 have?<br /></div></div>

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<div><div class="latex-wrapper">If the difference between the circumference and radius of a circle is 37 cm, using π = 22/7, what is the circumference (in cm) of the circle?<br /></div></div>

Question 9.

<div><div class="latex-wrapper">Let A and B be two events such that P(A)=0.25, P(B)=0.50 and P(A ∩ B)=0.14. The probability that neither A nor B occurs is:<br /></div></div>

Question 10.